Answer:
Sound waves are used for measurement in hertz.
Explanation:
Answer:
C₁₁H₁₂NO₄
Explanation:
In order to determine the empirical formula of doxycycline, we need to follow a series of steps.
Step 1: Determine the centesimal composition
C: 59.5 mg/100 mg × 100% = 59.5%
H: 5.40 mg/100 mg × 100% = 5.40%
N: 6.30 mg/100 mg × 100% = 6.30%
O: 28.8 mg/100 mg × 100% = 28.8%
Step 2: Divide each percentage by the atomic mass of the element
C: 59.5 /12.0 = 4.96
H: 5.40/1.00 = 5.40
N: 6.30/14.0 = 0.450
O: 28.8/16.0 = 1.80
Step 3: Divide all the numbers by the smallest one
C: 4.96/0.450 = 11
H: 5.40/0.450 = 12
N: 0.450/0.450 = 1
O: 1.80/0.450 = 4
The empirical formula of doxycycline is C₁₁H₁₂NO₄
Answer:
a bird sitting on a branch
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;

= 0.8
The rate-out
= 
= 
We can say that:

where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12

Integration of the above linear equation =

so we have:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places